Maximizing and minimizing two functions simultaneously Find $f$ which will maximize $y_1$ and minimize $y_2$:
$y_1 = \dfrac{ I f}{ 1+a f} $
$y_2 = (bf^3 + cf)\dfrac{I}{y_1}$
Note : $I , a, b , c > 0  ; f > 0 $
Thanks. 
 A: Generally when we have two functions and we want to maximize one and simultaneously minimize the other if we do it independently, generally we would get different $f$'s.Therefore we should assign a weight to each of them, according to their priorities. One way is to minimize (maximize) their difference, another one is minimizing one plus reciprocal of the other, etc. So generally there is no a unique answer to this question. Even saying considering trade-off is a qualitative statement and it could have lots of quantitative interpretations. 
When we do a trade-off, the $f^*$ would not be global extremum of both of them anymore. 
Consider we are maximizing $y_1$ and minimizing $y_2$ simultaneously where:
$$
y_1 = e^{-(x-1)^2}, \qquad y_2 = 2 - e^{-(x-2)^2}
$$
which are plotted here :

Now consider these two functions:
$$
y_3 = y_1 - y_2, \qquad y_4 = y_1 + \frac1{y_2}
$$
Which are plotted here :

It is easy to see that maximums of each of them occur in different places. 
Also note that generally it is not possible to find an $f^*$ satisfying $y_1(f^*)<y_1(f)$ and $y_2(f^*)>y_2(f)$ simultaneously.
